Tìm các số nguyên x và y biết :
a) \(5.2^{2x+1}.2^{-2}-2^x=384\)
b) \(3^{x+2}.5^y=45^x\)
c) \(\left(x+1\right)^{x+1}=\left(x+1\right)^{x+1}\)
d) \(27< 3^x< 243\)
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a) \(5.2^{x+1}.2^{-2}-2^x=384\Leftrightarrow2^x\left(5.2^{-2}.2-1\right)=384\)\(\Leftrightarrow2^x.1,5=384\Leftrightarrow2^x=384:1,5=256=2^8\)
\(\Rightarrow x=8\)
b) \(3^{x+2}.5^y=45^x\Leftrightarrow3^{x+2}.5^y=3^{2x}.5^x\Leftrightarrow\frac{3^{2x}}{3^{x+2}}=\frac{5^y}{5^x}\)\(\Leftrightarrow3^{2x-x+2}=5^{y-x}\Leftrightarrow3^{x+2}=5^{y-x}\)
\(\Rightarrow x+2=y-x=0\Rightarrow x=y=-2\)
a) \(5.2^{x+1}.2^{-2}-2^x=384\)
\(\Leftrightarrow2^x.2.\frac{5}{4}-2^x=384\)
\(\Leftrightarrow2^x.\left(\frac{5}{2}-1\right)=384\)
\(\Leftrightarrow2^x.\frac{3}{2}=384\)
\(\Leftrightarrow2^x=256\)
\(\Leftrightarrow2^x=2^8\)
\(\Leftrightarrow x=8\)
c) \(\left(x+1\right)^{x+1}=\left(x+1\right)^{x+3}\)
\(\Leftrightarrow\left(x+1\right)^{x+3}-\left(x+1\right)^{x+1}=0\)
\(\Leftrightarrow\left(x+1\right)^{x+1}\left[\left(x+1\right)^2-1\right]=0\)
\(\Leftrightarrow\orbr{\begin{cases}\left(x+1\right)^{x+1}=0\\\left(x+1\right)^2-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x+1=0\\\left(x+1\right)^2=1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\x\in\left\{0;-2\right\}\end{cases}}}\)
Vậy \(x\in\left\{0;-1;-2\right\}\)
a)Áp dụng bđt \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) ta có:
\(\left|x-1\right|+\left|3+x\right|=\left|1-x\right|+\left|3+x\right|\ge\left|1-x+3+x\right|=4\)
\(\Rightarrow VT\ge VP."="\Leftrightarrow-3\le x\le1\)
b) \(\hept{\begin{cases}\left|2x+3\right|+\left|2x-1\right|=\left|2x+3\right|+\left|1-2x\right|\ge4\\\frac{8}{2\left(y-5\right)^2+2}\le4\end{cases}}\Leftrightarrow VT\ge VP."="\Leftrightarrow\hept{\begin{cases}-\frac{3}{2}\le x\le\frac{1}{2}\\y=5\end{cases}}\)
c Tương tự b
2) \(\frac{1}{x}+\frac{1}{y}=5\Leftrightarrow x+y-5xy=0\Leftrightarrow5x+5y-25xy=0\Leftrightarrow5x\left(1-5y\right)-\left(1-5y\right)=-1\)
\(\Leftrightarrow\left(5x-1\right)\left(1-5y\right)=-1\)
Xét ước
Bài 3:
a: \(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\\dfrac{3}{4}x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{4}{3}\end{matrix}\right.\)
b: \(\Leftrightarrow\left\{{}\begin{matrix}3x+2>0\\\dfrac{2}{3}x-5< 0\end{matrix}\right.\Leftrightarrow-\dfrac{2}{3}< x< \dfrac{15}{2}\)
c: \(\Leftrightarrow\left[{}\begin{matrix}\dfrac{3}{4}x+2=0\\\dfrac{2}{5}x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\cdot\dfrac{3}{4}=-2\\\dfrac{2}{5}x=6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{8}{3}\\x=6:\dfrac{2}{5}=15\end{matrix}\right.\)
d.
33 < 3x < 35
--> 3 < x < 5
suy ra x=4